Why Convolution Preserves Integrability
Convolution combines two functions by averaging their overlap across translations. Even when \(f\) and \(g\) are each integrable, the formula for their convolution involves an integral depending on a new variable \(x\), so its existence and integrability need justification. Tonelli’s theorem supplies both: it controls the integral of the absolute value over the \((x,y)\)-plane and shows that the defining integral is absolutely convergent for almost every \(x\).
All measures below are Lebesgue measure \(m\) on \(\mathbb{R}\), and all functions \(f,g\) are measurable, complex-valued functions in \(L^1(\mathbb{R})\). Thus \(\|f\|_1=\int_{\mathbb{R}}|f(x)|\,dx<\infty\), and functions that agree \(m\)-a.e. represent the same element of \(L^1(\mathbb{R})\). The measure space \((\mathbb{R},\mathcal{L},m)\) is complete and sigma-finite.
To handle measurability directly, choose Borel measurable representatives of \(f\) and \(g\). Such representatives exist for Lebesgue measurable functions, up to changes on null sets. The map \((x,y)\mapsto f(x-y)g(y)\) is then Borel measurable on \(\mathbb{R}^2\). Changing either representative on a null set does not change the integral defining the convolution for any fixed \(x\): a change in \(g(y)\) occurs only on a null set of \(y\), and a change in \(f(x-y)\) occurs only when \(y\) belongs to a translate of a null set. Thus the resulting convolution is independent of the chosen representatives wherever its defining integral exists.
The Absolute Integral and Almost-Everywhere Existence
Proof. The function \((x,y)\mapsto |f(x-y)g(y)|\) is nonnegative and measurable on the product Lebesgue space. Tonelli’s theorem applies to the sigma-finite measure spaces \((\mathbb{R},\mathcal{L},m)\) in both variables. Its parameter-integral conclusion gives measurability of the integral in \(y\), and Tonelli also gives
For each fixed \(y\), translation invariance of Lebesgue measure, with the change of variable \(u=x-y\), gives \(\int_{\mathbb{R}}|f(x-y)|\,dx=\int_{\mathbb{R}}|f(u)|\,du=\|f\|_1\). Consequently, the displayed double integral equals \(\|f\|_1\int_{\mathbb{R}}|g(y)|\,dy=\|f\|_1\|g\|_1<\infty\). A nonnegative measurable function with finite integral is finite a.e.: indeed, if it were infinite on a set of positive measure, its integral would be infinite. Therefore the inner absolute integral is finite for \(m\)-a.e. \(x\), as claimed. \(\square\)
The \(L^1\) Convolution Inequality
Proof. The proposition proves absolute convergence a.e. The convolution is measurable: the positive and negative parts of the real and imaginary parts of the measurable integrand have measurable parameter integrals by Tonelli’s theorem; on the measurable set where the absolute integral is finite, their differences give the real and imaginary parts of \(f*g\). Extending by zero on the exceptional null set preserves measurability.
At every \(x\) where the defining integral is absolutely convergent, the triangle inequality for the Lebesgue integral gives
Both sides are nonnegative measurable functions, and the inequality holds a.e. Integrating it and applying Tonelli’s theorem to the right-hand side, as in the proposition, yields \(\int_{\mathbb{R}}|f*g(x)|\,dx\le\|f\|_1\|g\|_1<\infty\). Hence \(f*g\in L^1(\mathbb{R})\) and the asserted norm inequality follows. If either norm is zero, the same argument gives \(\|f*g\|_1=0\), so the conclusion also covers these degenerate cases. \(\square\)
Worked Example: Two Nonnegative Interval Indicators
Take \(f=\chi_{[0,1]}\) and \(g=\chi_{[0,2]}\) on \(\mathbb{R}\). Endpoints are included in these definitions, but their values at endpoints do not affect any of the integrals. For each \(x\), the convolution is the length of the set of \(y\) satisfying both \(0\le y\le2\) and \(0\le x-y\le1\). Equivalently, \(y\in[0,2]\cap[x-1,x]\). This set has length \(x\) for \(0\le x\le1\), length \(1\) for \(1\le x\le2\), and length \(3-x\) for \(2\le x\le3\); it has length zero outside \([0,3]\). Thus, up to endpoint values,
Since \(f*g\ge0\), its \(L^1\) norm is the sum of the three areas: \(\int_0^1x\,dx=\tfrac12\), \(\int_1^2 1\,dx=1\), and \(\int_2^3(3-x)\,dx=\tfrac12\). Therefore \(\|f*g\|_1=\tfrac12+1+\tfrac12=2\). Also, \(\|f\|_1=\int_0^1 1\,dx=1\) and \(\|g\|_1=\int_0^2 1\,dx=2\), so \(\|f\|_1\|g\|_1=1\cdot2=2\). Check: the direct area calculation gives \(2\), exactly the bound \(2\).
Worked Example: Cancellation Makes the Inequality Strict
Let \(f=\chi_{[0,1)}-\chi_{[1,2)}\) and \(g=\chi_{[0,1)}\). The half-open intervals specify the signs at the shared endpoint; changing those endpoint values would not change the integrals. For fixed \(x\), the convolution is the length of the \(y\)-values in \([0,1)\) for which \(x-y\in[0,1)\), minus the length for which \(x-y\in[1,2)\). Intersecting these intervals gives, up to endpoint values,
The first piece contributes \(\int_0^1x\,dx=\tfrac12\) to the norm. On \([1,2]\), \(3-2x\) is nonnegative for \(1\le x\le\tfrac32\) and nonpositive for \(\tfrac32\le x\le2\). The two absolute areas are \(\int_1^{3/2}(3-2x)\,dx=\tfrac14\) and \(\int_{3/2}^2(2x-3)\,dx=\tfrac14\). The last piece contributes \(\int_2^3(3-x)\,dx=\tfrac12\). Hence \(\|f*g\|_1=\tfrac12+\tfrac14+\tfrac14+\tfrac12=\tfrac32\). Meanwhile, \(\|f\|_1=1+1=2\) and \(\|g\|_1=1\), so \(\|f\|_1\|g\|_1=2\cdot1=2\). Check: \(\tfrac32\le2\), and the inequality is strict.
Worked Example: An Interval Indicator and a One-Sided Exponential
Set \(f=\chi_{[0,1]}\) and \(g(y)=e^{-y}\chi_{[0,\infty)}(y)\). These are nonnegative, so the convolution is nonnegative. The condition \(x-y\in[0,1]\), together with \(y\ge0\), gives no contribution when \(x<0\), gives \(0\le y\le x\) when \(0\le x\le1\), and gives \(x-1\le y\le x\) when \(x\ge1\). Therefore
Its norm is the integral of this nonnegative function. The first nonzero interval contributes \(\int_0^1(1-e^{-x})\,dx=e^{-1}\). For \(x\ge1\), \(e^{-(x-1)}-e^{-x}=(e-1)e^{-x}\), so the remaining contribution is \((e-1)\int_1^\infty e^{-x}\,dx=(e-1)e^{-1}=1-e^{-1}\). Thus \(\|f*g\|_1=e^{-1}+1-e^{-1}=1\). Directly, \(\|f\|_1=1\) and \(\|g\|_1=\int_0^\infty e^{-y}\,dy=1\), so the product of the norms is \(1\cdot1=1\). Check: the explicit convolution has norm \(1\), equal to the bound \(1\).
What the Bound Does—and Does Not—Say
The proof depends on the absolute-value double integral, not on cancellation in the signed or complex-valued integrand. Tonelli’s theorem is applicable to the nonnegative measurable function \((x,y)\mapsto|f(x-y)||g(y)|\); its finite integral is exactly \(\|f\|_1\|g\|_1\). This is what ensures almost-everywhere existence and gives an integrable upper bound for \(|f*g|\). No pointwise boundedness of \(f\) or \(g\) is required.
A common mistake is to discard the absolute values and use cancellation to estimate an \(L^1\) norm. In the signed example, direct integration of the displayed convolution gives \(\int_0^1x\,dx=\tfrac12\), \(\int_1^2(3-2x)\,dx=0\), and \(\int_2^3-(3-x)\,dx=-\tfrac12\), whose sum is \(0\). That calculation is not the \(L^1\) norm: the norm is \(\int|f*g|\,dx=\tfrac32\), as computed by splitting at \(x=\tfrac32\). The inequality concerns this absolute-value integral.
The conclusion is a norm estimate for the convolution itself; it does not assert that equality always holds. The nonnegative examples attain equality, while the signed example shows that cancellation can make the inequality strict. The essential general strategy is to establish absolute convergence almost everywhere first and then integrate the pointwise triangle inequality, with Tonelli’s theorem justifying the double integral.
Check Your Understanding
For each question, keep track of whether the claim is pointwise or only \(m\)-a.e., and identify the role of Tonelli’s theorem.
- For \(f,g\in L^1(\mathbb{R})\), write the absolute-value double integral used to prove that \(f*g\) exists absolutely a.e.
- Why does finiteness of \(\int_{\mathbb{R}}\int_{\mathbb{R}}|f(x-y)||g(y)|\,dy\,dx\) imply that the inner integral is finite for a.e. \(x\)?
- For \(f=\chi_{[0,1]}\) and \(g=\chi_{[0,2]}\), identify the three intervals on which the convolution has a nonzero formula, and verify its \(L^1\) norm.
- In the signed example, why does computing the signed integral of \(f*g\) not compute \(\|f*g\|_1\)?
- If \(\|f\|_1=0\), what does the convolution inequality imply about \(\|f*g\|_1\), and what does that mean for the convolution as an \(L^1\) element?